Quantized redshift anomaly

This is from an interesting website :Galilean Electrodynamics. It's a pay site, but I did find some free stuff. Here is a sample article they posted, a very good explanation of what light does. I excerpted something of interest to me which I copied below. If we see something in the light of day, what are we seeing?

mywebpages.comcast.net/adring/Persson_invisible.pdf

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">



<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The Ether

Electric, magnetic and gravitational forces are physical realities,
and so are electromagnetic waves. The owner of these properties
is something real that we call ether (unfortunately denied).
The ether does the ‘waving’. The existence of an ether means
that emission and absorprtion can be considered as electron/
ether interactions, rather than as electron/light interactions.
Therefore the energy absorbed by an electron is not necessarily
transferred by the light. The fact that, when switched on, light
affects electrons instantainously supports the idea of an electron/
ether interaction.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Conclusions
The black-body radiation law, the photoelectric effect,
Compton effect, and Crookes' radiometer can all be united with a
wave model of electromagnetic radiation without particle structure.
Light-quanta are illusions caused by the discreteness of electrons
and the invisibility of light.

Planck's constant provides information about an electron’s relation
to an atomic nucleus, but implies nothing about light. His
constant implies ‘quantizition’ in matter, and consequently in
emission as well as in absorption of light, but not during light
transmission.

The ether is a physical reality. Quantization is a mathematical
necessity and does not prove zero process time.
The constancy in light speed demonstrates this speed to be a
property of an ether.

Remarks

The fact that light is not quantized at the Planck level does not
exlude the possibility that the ether could be quantized at a lower
level. The ether could, for instance, be constituted of undetectable
neutrinos. A velocity vector could be defined as an average
value of many neutrino velocities, and thereby define the velocity
of an ether.

The fact that moving electrons demonstrate a wavelike behavior
in some experiments can also be explained by an
ether/electron interaction. A ship moving in water can serve as a
metaphor for this idea.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Tommy</i>
<br />I often wonder what light is. They talk of photons, and then waves...must be a special kind of wave that doesn't wave outwards...and what kind of particle goes on forever...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Light has all known wave properties and no essential particle properties. The two non-particle properties, the Compton effect and the Photoelectric effect, both have wave interpretations also. So what makes most sense is that light is a pure wave rather than some sort of poorly defined particle-wave hybrid.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">What, in terms of the redshift anomaly, is the relationship between the color of light (frequency) and spectral lines?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Color has a one-to-one relation to frequency. Spectral lines are caused by the absorption (or emission) of particular frequencies, usually in a very narrow range. Such lines correspond to the elements or compounds that produced them. For example, hydrogen has its own sets of unique lines at certain laboratory frequencies.

A redshift means that the spectral lines are shifted to the red side of those basic laboratory frequencies. It generally says nothing about the color of the object.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Question, isn't our problem about spectral line shifting? And if it is about absorbtion changes, then isn't it the change of absorbtion (frequency) that matters, and not light itself? So that instead of talking about the energy of a photon, we should be talking about the energy (effects) of an electron field?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Light is not a form of matter, but electrons are. Redshift is all about light, not electrons. Because redshift results from energy loss, it must occur after the light leaves an atom, assuming that atoms of a particular type are the same everywhere.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">what if it is the matter at that time that is different? Isn't that what we are actually observing? The effect of matter on light, not light itself.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">No, unless you mean only that light entering our eyes must interact with the matter in our eyes.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">(Actually light is invisible, what we see is the effect of light on matter.)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">What we "see" is light, unless you change definitions of the words. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">quote:

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Question, isn't our problem about spectral line shifting? And if it is about absorbtion changes, then isn't it the change of absorbtion (frequency) that matters, and not light itself? So that instead of talking about the energy of a photon, we should be talking about the energy (effects) of an electron field?

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Light is not a form of matter, but electrons are. Redshift is all about light, not electrons. Because redshift results from energy loss, it must occur after the light leaves an atom, assuming that atoms of a particular type are the same everywhere.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

What about as it makes its way THROUGH the atom? Is it correct to say that a photon, a packet of waves, does not make the 13 billion year journey to here without hitting something on the way. Is it correct to say that when a photon does nit something in the way, that something will emit a photon and the journey goes on. I think that is what I am reading. Oh, without scattering. So if that is so, then what we are looking at is the behavoir of matter, and not the light itself. It is the matter which takes in the photon and processes it outward again. That it would do this without energy loss is amazing...

The sample article reads..."An exponential dependency
in Wien's (not Planck's) law is identified as a sign of
entropy of radiation. Perhaps a better interpretation is entropy of
matter (electrons), observable in radiation. We must remember that
observation of light involves absorption by atoms. .Variations of
black-body radiation with higher frequencies could emanate
from the behavior of the matter rather than of the light itself."

What he is saying is that it is the matter that is affecting the light and not the light that is being affected. Light does what it is supposed to be doing, matter is where the energy loss is to be located.

Keep in mind that I have no idea what I am talking about, I don't even know what light is. But if one looks at light, clearly we are not seeing the light, we are seeing what stopped the light. Light itself is invisible. (I think that is because light is so small)And the point I am making perhaps niavely, is that since light is interacting with amtter in order to be seen, maybe it is the matter that we should be looking at.

So what is it about the matter in deep space that would affect light differently? Less efficient then more efficient now. More friction then, less friction now. Perhaps "then" has nothing to do with it. Perhaps the energy anomaly is a result of the atomic process going on. Perhaps a quasar is an emergence of energy from or through the Aether, and when energy, to be more precise, when gamma rays come close to an ion, it can create electron/positron pairs, perhaps this first emergence of matter is reflective of low energy photons. So instead of photons losing energy over time, photons actually, according to this scenerio, gain energy? Does that make any sense?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Tommy</i>
<br />What about as it makes its way THROUGH the atom?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Nope. Lightwaves are either propagating through vacuum or absorbed by matter.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Is it correct to say that a photon, a packet of waves, does not make the 13 billion year journey to here without hitting something on the way.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">First, a "photon" is a hypothetical wave packet, a single entity, not multiple entities as in your "packet of waves". Each wave crest registers in detectors as another "photon".

And light mostly does travel the 13 billion lightyears to here without hitting anything. Photons that do hit hydrogen or other intergalactic atoms are absorbed, and show up here as spectral lines (missing photons). If photons did interact with oedinary matter, they would be slightly scattered, and images of distant galaxies would be fuzzy. But Hubble observations do not show this.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Is it correct to say that when a photon does hit something in the way, that something will emit a photon and the journey goes on.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The two possible fates are absorption or scattering. Most photons suffer neither, so they have not interacted with matter.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Light does what it is supposed to be doing, matter is where the energy loss is to be located.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Redshift (energy loss) mechanisms are many. The main mechanism in BB (Doppler) and that in MM (friction with gravitons) do not involve any interaction with ordinary matter. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
Light has all known wave properties and no essential particle properties.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I completely agree. Unhappily, winning his fight against Planck through his first Nobel prize, Einstein maide his worst (almost unique in Physics) error. For Planck, the quantization is a property of matter.

The main arguments against Einstein's quantization of light are:

- In its usual form E=hf (f: frequency), Planck's law is absurd because a pure frequency which requires an infinite sin function has no physical meaning. It is necessary to introduce a bandwidth, or, worse a shape of line to make Planck's formula meaningful.
Consequently, a photon would result, at least on two parameters (mean frequency, linewidth), better on many parameters. Too many photons !

- A lot of people made experiments to show that quantum electrodynamics is better than classical electrodynamics. The experiments are good, but their classical explanations are bad. The errors come from a lack of knowledge of the theory of the modes (developped in the 19th century, in acoustics); Maxwell's equations in the vacuum (and with sources replaced by their advanced fields, and linear conditions at the limits) are linear, that is any linear combination of solutions, with constant, real coefficients, is a solution; in modern language, the solutions build a real, generally infinite space. The modes are "rays" of this space, that is sets of proportional solutions. Any of these solutions depends, therefore on a <b>single real number</b>, the amplitude of the solution (do not confuse this amplitude with the amplitudes of the field in each point). The absorption of a field requires the addition of a field having an opposite amplitude, in the same mode. Generating this opposite field is tried in acoustics (active absorption of a noise), but, evidently it does not work well. It is not a big problem in acoustics because there are attenuations( which are not present in Maxwell's equations). Thus, in electrodynamics, the amplitudes in the modes have a minimal value corresponding to the zéro point field.
An emission by a source is an amplification of the field in the mode characteristic of the source (spontaneous emission if it reduces to the ZPF), an absorption is a decrease of the absolute walue of the amplitude, at the best down to the ZPF. The ZPF is a thermodynamical bath: If the energy in a mode is increased by the emission of hf by an atom, the amplitudes in non-orthogonal modes corresponding to possible emissions of other atoms increase (in the average), so that these atoms may absorb hf. Consequently, the emission of hf is followed by an absorption of 0, hf, 2hf , 3hf... by 0, 1 ,2 ,3 ... atoms; in the average only of hf.
If a photoelectric cell receiving an amplitude E absorbs it "completely", it leaves a ZPF F, so that it receives an energy proportional to W=EE-FF=(E-F)(E+F). If E&gt;&gt;F, as usual, w~EE, but, in "photon counting", where F&gt;&gt;E, W~F(E-F), proportional to the field, not to its square. The promoters of QED do not want to use the good formula in their classical computations !

- The emission of a quantum is stimulated by the field in its mode. If we use a plane wave to excitate a dipole, this field must be split into a "spherical" field and diffracted fields. Thus, only a fraction of the electric field at the dipole excitates it. On the contrary, the spontaneous emission is produced by the ZPF in the spherical mode: no need to reduce the effective field. The experiments verifies this.
In QED, the incident field is not split; to get the experimental result, it is necessary to introduce a strange, ad hoc "radiation reaction"

It is important to note that there are two types of light-matter interactions. Usually, one considers interactions in which the state of the matter changes, so that the change of the energy of the wave(s) is quantified. In "parametric interactions", matter excited by the electromagnetic field returns to its stationary state after the interaction, so that there is no quantization.
The simplest parametric interaction is refraction in which a very low light energy may excite a big prism (temporarily)! Other interactions allow exchanges of energy between light beams simultaneously refracted by a convenient medium: one obtains frequency combinations, doubling, splitting in laser technology. In astrophysics, we are particularily interested by the transfers of energy which allow a redshift of light, and a simultaneous blueshift of the radio waves (CREIL effect); but the convenient matter is uncommon using ordinary light: excited atomic hydrogen.

Replying to JMB:

I mostly could not understand what you said. However, it seems relevant (not to mention simpler) to point out that "quantization" for light simply means that light consists of discrete waves instead of some sort of continuum. And each wave at a given frequency has a specific energy, so the total energy transferred by light to matter must come in integer multiples of the energy transferred from a single wave. -|Tom|-